Inductive reactance and length of the solenoid

[Anthem26]

Homework Statement

A solenoid with a radius of 8.0 10-3 m and 170 turns/cm is used as an inductor in a circuit. When the solenoid is connected to a source of 15 V rms at 22 kHz, an rms current of 3.7 10-2 A is measured. Assume the resistance of the solenoid is negligible.
(a) What is the inductive reactance?
(b) What is the length of the solenoid?

r = 8E-3 m
n = 170E2 turns/m
Vrms = 15 V
f = 22E3 Hz
Irms = 3.7E-2 A
Area = pir^2 = 2.01E-4 m^2

Homework Equations

XL = 2pifL

L = μn^2pir^2l

The Attempt at a Solution

I know how to utilize the equations, but my problem is finding L (self-inductance) in order to solve both part a and b. I try looking throughout my physics book, but nothing.

a) XL = 2*pi*22E3*L

b) L = 4piE-7*17000^2*3.14*8E-3*l

 

[Anthem26]

Nevermind, I managed to figure the problem out myself.

 

[kerimek]

To solve for L, we need to use the inductance formula L = μn^2πr^2l, where μ is the permeability of free space, n is the number of turns per unit length, r is the radius of the solenoid, and l is the length of the solenoid.

We can rearrange the formula to solve for L:

L = XL/(2*pi*f)

Substituting the given values, we get:

L = (3.7E-2 A)/(2*pi*22E3 Hz) = 2.67E-6 H

To find the length of the solenoid, we can use the formula:

l = L/(μn^2*pi*r^2)

Substituting the known values, we get:

l = (2.67E-6 H)/(4piE-7 *170^2/m * pi * (8E-3 m)^2) = 0.095 m

Therefore, the inductive reactance is 2.67E-6 ohms and the length of the solenoid is 0.095 meters.